Let X represent initial speed. Then speed at point A is X/3 and speed at B is X/4.
Distance between A and B is equal to:
@$Dab = ((X/3)^2-(X/4)^2)/(2*g) = 28@$
Solving the equation for X, obtaining X = 106.3 m/s.
Maximum height is determined as follows.
@$Hmax=X^2/(2*g) = 106.3^2/(2*9.81) = 576 m@$